# Simple software low pass filter

I am measuring voltages and currents of a 3-phase electrical machine and I need to calculate the power. Every interrupt (frequency between 30-70kHz) I get values (voltages, currents) from analogue-digital converter and I need to do a simple calculation to determine power. However, I was told to use averaging with a first order filter with a time constant of approximate 1s for the power calculation. This, I believe, means low pass filter with a frequency of approximately 1Hz? What is the simplest solution to this? Moving average?

A first order lowpass filter is usually implemented like this:

$$p[n] = \alpha p[n-1] + (1-\alpha) pi[n]$$

Where $p[n]$ is your filtered power estimation, $p[n-1]$ is the previous result, $pi[n]$ is your new measurement (probably the product of instantaneous voltage and current measurements), and $\alpha$ is a positive parameter just less than 1.

The nearer $\alpha$ gets to 1, the larger the time constant (lower cutoff frequency) of your filter. But beware, especially in embedded systems with limited precision, that getting too near to 1 can make your filter unstable, or at least have problems due to numerical precision.

The cutoff frequency for that filter is around $f_s \frac{1-\alpha}{2\pi\alpha}$, where $f_s$ is your sampling frequency.

• There must be a mistake in the formula for cutoff freq ($\alpha/(1-\alpha)$ is much larger than 1 if $\alpha$ is close to but less than 1, no? I think it is $\frac{1-\alpha}{2\pi \alpha}$ – Dan Boschen Apr 9 '17 at 0:03
• Thanks, this looks similar to the solution at the bottom of this page: norwegiancreations.com/2015/10/… – jurij Apr 9 '17 at 16:12
• However, your equation states that current measured value needs to be multiplied by (1-alpha) and previous value with alpha, while on the page above, this is vice versa. Which is correct? Also, which equation for cutoff frequency is correct? Thank you very much. – jurij Apr 9 '17 at 16:14
• $\alpha$ happens to be the pole location of this simple LPF. i would recommend that $\alpha$ be very close to 1, but less than 1. $$0 < 1-\alpha \ll 1$$. and $pi[n]$ is the instantaneous power at time $n$ and is the product of the instantaneous voltage and instantaneous current as measured by the separate A/D converters. this is necessary to measure real power in a reactive situation. – robert bristow-johnson Apr 9 '17 at 17:59
• OK, but in this case, do I have to multiply pi[n] with alpha or with (1 - alpha)? Which is the correct equation for cutoff frequency? – jurij Apr 9 '17 at 21:04